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JIPMER 2006
Exam
JIPMER
Year
2006
Subject
Physics
Chapter
Motion in One Dimension

JIPMER 2006 · Physics · Motion in One Dimension

A car starts from rest, moves with an acceleration a and then...

A car starts from rest, moves with an acceleration aa and then decelerates at a constant rate bb for sometimes to come to rest. If the total time taken is tt. The maximum velocity of car is given by :

Options

  1. Aabt(a+b)\frac{a b t}{(a+b)}
  2. Ba2ta+b\frac{a^2 t}{a+b}
  3. Cat(a+b)\frac{a t}{(a+b)}
  4. Db2ta+b\frac{b^2 t}{a+b}

Correct answer

A. abt(a+b)\frac{a b t}{(a+b)}

Step-by-step solution

Let car acceleates for time t1t_1 and decelerates for time t2t_2 then t1+t2=tt_1+t_2=t.....(1) From v=u+a t v=u+at1⇒v=at1\begin{aligned} & v =u+a t_1 \\ \Rightarrow v & =a t_1\end{aligned} For deceleration v=u−at0=at1−bt2(∵u=v)at1=bt2⇒t2=at1b∴t1+at1b=1[FromEq.(1)]⇒t1(1+ab)=t⇒t1=bta+b\begin{aligned} & v=u-a t \\ & 0=a t_1-b t_2 (\because u=v)\\ & a t_1=b t_2 \\ & \Rightarrow \quad t_2=\frac{a t_1}{b} \\ & \therefore \quad t_1+\frac{a t_1}{b}=1 [From Eq. (1)]\\ & \Rightarrow \quad t_1\left(1+\frac{a}{b}\right)=t \\ & \Rightarrow \quad t_1=\frac{b t}{a+b} \\ & \end{aligned}\thereforeMaximumvelocityofcarMaximum velocity of carv=a t_1=\frac{a b t}{a+b}$
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